在斜三角形ABC中tanC/tanA+tanC/tanB=1,则（a^2+b^2)/c^2
问题描述：
在斜三角形ABC中tanC/tanA+tanC/tanB=1,则（a^2+b^2)/c^2
答
tanC/tanA+tanC/tanB=1
tanBtanC+tanAtanC=tanAtanB
tanC(tanA+tanB)=tanAtanB
sinC/cosC (sinA/cosA+sinB/cosB)=sinAsinB/cosAcosB
sinC (sinAcosB+cosAsinB)=sinAsinBcosC
sinC sin(A+B)=sinAsinBcosC
sinC sin(180°-C)=sinAsinBcosC
sin^2C=sinAsinBcosC
sinC/sinA * sinC/sinB = cosC
c/a * c/b = (a^2+b^2-c^2)/(2ab)
2c^2 = a^2+b^2-c^2
a^2+b^2 = 3c^2
(a^2+b^2)/c^2 = 3